It's a long time since I had a problem compiling R, but I've now
encountered one with R-2.6.2 on Fedora Core 6 (64bit).
The problem arises when the installation process gets to the grid
package, the unit.c part in particular. This is the output I get:
gcc -std=gnu99 -I../../../../include -I../.
It's your system: try unpacking the tarball again. That line is
eval(R_fcall3, R_gridEvalEnv);
and has no # there.
I suspect FC6 systems are thin on the ground (it is end-of-life, I
believe), but R-2.6.2.tar.gz has been installed on many, many 64-bit Linux
systems (e.g. CRAN test
dear list,
as a part my problem. I have to estimate some parameters using ML
estimation. The form of the likelihood function
is not straight forward and I had to use a for loop to define the function.
I used "optim" to maximise the result but
was not sure of the programme.
To validate my results, I
Hello,
I have a matrix with 4000 rows and 8 columns with 1 and zeros
Every 1000 rows comes for a different rule, f1,f2,f3,f4.
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
...
1 1 0 0 1 1 0 0
1 1 0 0 1 1 0 0
...
0 0 1 1 0 0 1 1
0 0 1 1 0 0 1 1
How can I extract 4 matri
Hi
I have some problems in defining new generic functions and classes. Just
have a look at the following example:
require(fPortfolio)
setClass("PROBECLASS",
representation(
type="character"
)
)
isGeneric("setType
It works! I was banging my head for a week.
thank you so much.
Bill.Venables wrote:
>
> One way to do it is to find the distances between ther "centers" (=
> centres in English) of the clusters.
>
> dist(kc$centers)
>
> It rather depends on how you define distances between clusters, thou
The new package obsSens is now on the CRAN mirrors. This package has tools for
doing senstitivity analysis for observational studies.
The common criticism of observational studies is that there is the possibility
of an unmeasured variable that is related to both the response and the
predictor
First of all thank you for the responses. I appreciate the
suggestions i have received thus far.
Just to reiterate
I am trying to analyze a data set that has been collected from a
hierarchical sampling design. The model should be a mixed model
nested ANOVA. The purpose of my study is to analyz
Hi all,
Can anyone who is familar with CART tell me what I missed in my tree code?
library (MASS)
myfit <- tree (y ~ x1 + x2 + x3 + x4 )
# tree.screens () # useless
plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0)
# tile.tree (myfit, fgl$type) # useless
# cl
> /I think your question should be more relevant on Rdev./
ok, I will
> Personnally I would find stuff like "names", "$", "$<-", or "["
> useful as these are usual operation with S3 objects.
Is it possible in S4 to define "$<-" ? If there is a slot name 'a' in
object 'B', I find (in "S4 in 15 p
Christophe Genolini wrote:
>> /I think your question should be more relevant on Rdev./
> ok, I will
>> Personnally I would find stuff like "names", "$", "$<-", or "["
>> useful as these are usual operation with S3 objects.
> Is it possible in S4 to define "$<-" ? If there is a slot name 'a' in
See the 'useAsDefault' argument to setGeneric.
As an aside, if 'setType<-' is meant to be a 'setter' to change the
value of a slot 'type', then I find the syntax a little redundant --
it's use
> setType(x) <- "foo"
implies that it is already a 'setter' without 'set' at the front. Why
not jus
Hello,
Is there a way to add minor tick marks to the Y-axis of a lattice
plots?
Thank you
Saptarshi
Saptarshi Guha | [EMAIL PROTECTED] | http://www.stat.purdue.edu/~sguha
[[alternative HTML version deleted]]
__
R-hel
Martin Morgan wrote:
> Christophe Genolini wrote:
>
>>> Personnally I would find stuff like "names", "$", "$<-", or "["
>>> useful as these are usual operation with S3 objects.
>>
>> Is it possible in S4 to define "$<-" ? If there is a slot name 'a' in
>
>
> Possible yes, see below. Usual? It i
Hi all,
As R is gaining space in our research group as language and environment
for statistical computing and graphics it will be really helpful to have
the option of using R as a web service.
We are currently using a collaborative edition wiki tool (XWiki) to
develop R code. And we can call t
Hi Everybody,
A new package called ``Bayesian Prediction with High-order
Interactions'' is available from CRAN. The description of this package
is as follows"
"This R package is used in two situations. The first is to predict the
next outcome based on the previous states of a discrete sequence. T
o ha wang wrote:
> Hi all,
>
> Can anyone who is familar with CART tell me what I missed in my tree code?
>
> library (MASS)
> myfit <- tree (y ~ x1 + x2 + x3 + x4 )
There is not function tree() in MASS. I guess you have loaded package
"tree"?
Note that the author and maint
Evgenia wrote:
> Hello,
> I have a matrix with 4000 rows and 8 columns with 1 and zeros
> Every 1000 rows comes for a different rule, f1,f2,f3,f4.
>
> 1 1 1 1 0 0 0 0
> 1 1 1 1 0 0 0 0
>
> 0 0 0 0 1 1 1 1
> 0 0 0 0 1 1 1 1
> ...
> 1 1 0 0 1 1 0 0
> 1 1 0 0 1 1 0 0
> ...
> 0 0 1 1 0 0 1 1
Hi,I am wondering how to conduct Kenward-Roger correction in
the linear mixed model using R. Any idea?
Thanks a lot,
Suyan
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Pedro Mardones wrote:
> Dear all;
> I'm kind of confused with the results obtained using the ginv function
> from package MASS and pinv function from Matlab. Accroding to the
> documentation both functions performs a Moore-Penrose generalized
> inverse of a matrix X. The problem is when I change
Michelle DePrenger-Levin wrote:
> Hello,
>
>
>
> I am trying to use the projection.matrix( ) function and am following the
> example given. I have my data formatted very similar to the test.census
> example.
>
>
>
>> str(AsMi05mat)
>
> `data.frame': 1854 obs. of 6 variables:
>
> $
stephen sefick wrote:
> I have two dataframes in R that were tab seperated .txt files
>
> y<-read.table("foo.txt", header=T)
> x<-read.table("foo.txt", header=T)
>
> these are set up like this:
>
> Datetime Temp
> 01/01/07 00:01 11.5
> 01/01/07 00:16 11.6
>
> e
On Sun, 24-Feb-2008 at 10:00AM +, Prof Brian Ripley wrote:
|> It's your system: try unpacking the tarball again.
No surprises there: Brian is spot on again. Thank you Brian.
I'd not have thought of trying unpacking again. I thought there would
be no stochastic element to such a process.
I guess you don't have any degrees of freedom left ...
How much observations versus how much estimated parameters do you have
for estimating your model parameters?
Uwe Ligges
Marcelo Luiz de Laia wrote:
> Hi,
>
> I run a maanova analysis and found this message error:
>
> Error in ma.svd(X, 0
Is it possible to get something like rapply for a data.frame? For
example, if I do this:
> x <- list(a=mtcars, b=mtcars)
> str(rapply(x, nrow, classes="data.frame", how="replace"))
List of 2
$ a:List of 11
..$ mpg : num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
..$ cyl :
Hello all:
Just wondering if I can get advice on what kind of
bootstrapping I should use when using a regression
model to estimate juvenile fish passage data. I use
rotary screw traps to do fish mark-recapture trials
and the efficiency of every trial is added to the
graph generating a different R-s
On 2/24/2008 4:03 PM, Felipe Carrillo wrote:
> Hello all:
> Just wondering if I can get advice on what kind of
> bootstrapping I should use when using a regression
> model to estimate juvenile fish passage data. I use
> rotary screw traps to do fish mark-recapture trials
> and the efficiency of eve
It's in the 'limma' Bioconductor package.
Next time you can try
help.search("lmFit")
RSiteSearch("lmFit")
Gabor
On Sun, Feb 24, 2008 at 01:02:41PM -0800, Keizer_71 wrote:
>
> Hi Everyone,
>
> I am trying to use lmFit function; however, i cannot find it function
> anywhere.
>
> I have been try
Michael Hoffman wrote:
> Is it possible to get something like rapply for a data.frame?
On second thought, I can do something like this:
happly <- function(x, fun) {
if (class(x) == "list") {
lapply(x, happly, fun)
} else {
fun(x)
}
}
> happly(list(a=list(c=mtcars, d=Titanic),
thank you.
Chuck Cleland wrote:
>
> On 2/24/2008 4:02 PM, Keizer_71 wrote:
>> Hi Everyone,
>>
>> I am trying to use lmFit function; however, i cannot find it function
>> anywhere.
>>
>> I have been trying to find the function in Bioconductor and elsewhere. I
>> re-install bioconductor source
On 2/24/2008 4:02 PM, Keizer_71 wrote:
> Hi Everyone,
>
> I am trying to use lmFit function; however, i cannot find it function
> anywhere.
>
> I have been trying to find the function in Bioconductor and elsewhere. I
> re-install bioconductor source, update package and update R as well. no luck
>
Hi Stephen,
>> Also i have read in Quinn and Keough 2002, design and analysis of
>> experiments for
>> biologists, that a variance component analysis should only be conducted
>> after a rejection
>> of the null hypothesis of no variance at that level.
Once again the caveat: there are experts on
Hi Everyone,
I am trying to use lmFit function; however, i cannot find it function
anywhere.
I have been trying to find the function in Bioconductor and elsewhere. I
re-install bioconductor source, update package and update R as well. no luck
Is there a command in R where i can just type, and i
o ha wang wrote:
> Thanks!
>
> Yes, you were right. I loaded package 'tree'.
>
> I tried 'tree' and 'rpart' respectively, it looks like results for
> regression tree are same, and very similar for classification tree. (more
> biref using 'rpart' than using 'tree').
>
> But my
Hi Stephen,
Slip of the dactylus: lm() does not, of course, take a fixed=arg. So you
need
To recap:
mod.rand <- lme(fixed=y ~ x, random=~x|Site, data=...)
mod,fix <- lm(y ~ x, data=...) ## or
##mod,fix <- lm(formula=y ~ x, data=...)
Bye.
Mark Difford wrote:
>
> Hi Stephen,
>
>>> Also
Thanks!
Yes, you were right. I loaded package 'tree'.
I tried 'tree' and 'rpart' respectively, it looks like results for regression
tree are same, and very similar for classification tree. (more biref using
'rpart' than using 'tree').
But my problems are still not been solved us
Dear R-helpers,
(1) I can't figure out how to tell Dotplot that I want the colors of
the CI bars to be the same as the colors of the dots.
For example:
t2a.ci3 <- data.frame(est = c(7, 20, 75), lower = c(-9, 0.5, 42),
upper = c(22, 39, 109))
mypal <- c('skyblue3', 'mistyrose3')
condNames <- c
>> Also i have read in Quinn and Keough 2002, design and analysis of
>> experiments for
>> biologists, that a variance component analysis should only be conducted
>> after a rejection
>> of the null hypothesis of no variance at that level.
Hmmm...
This does rather assume that 'no significant resu
Hi,
I am just starting to use R for a graduate course, and I like how the
correlation matrix at
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=137
I did something similar by copying from the examples(pairs), but it
seems that I need to jitter the bottom panel... and I have no idea
Hi the list
Is it possible to define a method with a signature that will have more
argument than the generic method?
For exemple, print has only one argument, plot has two, can I do
something like :
setMethod("print",signature=c(x="numeric",y="character"),function(x,y,...)
..
setMethod("pl
Hi all,
I would like to color the area between two time-series. I tried it by
using the polygon() function but I keeps drawing lines between beginning
and end points.
Is there another more appropriate function or how could I close the
polygon at the end en the beginning of the time series (e.g
Hello,
I have a vector of 1,000,000 numbers and another vector of 1,000
divisors. What I'd like to do is to divide the first 1,000 numbers of
the first vector by the first divisor, then the next 1,000 by the second
divisor and so on. I came up with this, but I was wondering if there is
a more idio
x/rep(divs,each=1000)
cheers,
Rolf Turner
On 25/02/2008, at 2:36 PM, Andre Nathan wrote:
> Hello,
>
> I have a vector of 1,000,000 numbers and another vector of 1,000
> divisors. What I'd like to do is to divide the first 1,000 numbers of
> the first vector by the first
x <- x/rep(divs, each = 1000)
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary): +61 7 3826 7304
Mobile: +61 4 8819 4402
Home Phone: +61 7 3286 7700
mai
How about
x <- x / rep(divs, rep(1000, 1000))
?
Cheers,
Andrew
On Sun, Feb 24, 2008 at 10:36:23PM -0300, Andre Nathan wrote:
> Hello,
>
> I have a vector of 1,000,000 numbers and another vector of 1,000
> divisors. What I'd like to do is to divide the first 1,000 numbers of
> the first vecto
I think you have to change one statement in your program:
xx <- cbind(time(z[,1]),rev(time(z[,2])))
On Sun, Feb 24, 2008 at 7:44 PM, <[EMAIL PROTECTED]> wrote:
> Hi all,
>
>
>
> I would like to color the area between two time-series. I tried it by
> using the polygon() function but I keeps dr
I have a data frame which contains some valuable date information. But for a
few of the dates, the day information missing .
Viz:
> interesting.data$date
[1] "1/22/93" "1/22/93" "1/23/93" "1/00/93" "1/28/93" "1/31/93" "1/12/93"
i.e. for dates where the day info is missing, the "%d" part of the
On Mon, 2008-02-25 at 11:52 +1000, [EMAIL PROTECTED] wrote:
> x <- x/rep(divs, each = 1000)
Fantastic :)
Thanks Bill, Andrew and Rolf
Andre
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On Sun, 2008-02-24 at 23:26 -0300, Andre Nathan wrote:
> > gp(1, 2, 10)
> [1]1248 16 32 64 128 256 512 1024
>
Actually,
[1]1248 16 32 64 128 256 512
Andre
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https:
On Feb 24, 2008, at 9:36 PM, Andre Nathan wrote:
> On Sun, 2008-02-24 at 23:26 -0300, Andre Nathan wrote:
>>> gp(1, 2, 10)
>> [1]1248 16 32 64 128 256 512 1024
>>
>
> Actually,
>
> [1]1248 16 32 64 128 256 512
>
2^(0:9)
>
> Andre
Haris Skiadas
It really depends on what you want to do with them but one possibility
might be to represent them as chron dates and use a time of 0 for true dates
and noon for missing dates replacing the missing day with 01 or 15 or some
other day:
> library(chron)
> x <- c("01/00/05", "01/22/06")
> no.day <- re
On Sun, 2008-02-24 at 21:39 -0500, Charilaos Skiadas wrote:
> 2^(0:9)
I guess it's so simple that I'd never think of that...
Thanks!
Andre
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Hi
I'm pretty sure there's some built-in function to do the equivalent of
this, but I couldn't find one anywhere:
gp <- function(init, mult, n)
{
if (n == 1)
init
else
pg(c(init, init[length(init)] * mult), mult, n-1)
}
> gp(1, 2, 10)
[1]1248 16 32 64 128 25
In looking at this again here is a slight simplification. Its now
only one line:
> library(chron)
> x <- c("01/00/05", "01/22/06")
> as.chron(sub("/00/", "/15/", x)) + (regexpr("/00/", x) > 0) / 2
[1] (01/15/05 12:00:00) (01/22/06 00:00:00)
On Sun, Feb 24, 2008 at 9:45 PM, Gabor Grothendieck
<[
Michelle,
I would probably run a loop as well and save all matrices to a single list
(see hudsonia and calathea on working with lists of matrices).
First, run the example(test.census) to get the stage-fate data frame "trans"
and then run this code to save the matrices into a list "all".
years<-
useR's,
Does any one know if there is a size limitation on the data frames that can
be included in R packages. I have a data set in a text file that I would
like to include in a package I am building and it is 8.5 MB in size. Will
this be problematic? Is the process for including data sets in
Hello!
I am working with signals and a plot of several signals on the same
axes can get quite messy. With lines that are very fractured,
distinction by only the linestyle is not very clear. If I add symbols
to the plot however, there are so many symbols, that they overplot and
the whole plot is un
Try this:
pch <- c("+", ""); lwd <- 1
plot(1:10, type = "o", pch = pch, lwd = lwd)
legend("topleft", legend = "data", pch = pch, lwd = lwd)
On Sun, Feb 24, 2008 at 11:01 PM, Tribo Laboy <[EMAIL PROTECTED]> wrote:
> Hello!
>
> I am working with signals and a plot of several signals on the same
>
Hi,
I am using randomForests for a classification problem. The predict
function in the randomForest library, when asked to return the
probabilities, has precision of two digits after the decimal. I need
at least four digits of precision for the predicted probabilities. How
do I achieve this?
Than
Dear R users.
I have written my R code. It runs in the version
2.3.1, but not in the version 2.5.1. The error
message is shown below.
Error in .Fortran("DEPTH", u, v, as.integer(size),
as.single(x), as.single(y), :
Fortran symbol name "depth" not in load
table
Could an
Hi
I can not help you to run your code in R 2.5.1 but I would recommend to
use even more recent R version (2.6.1) or even maybe R 2.7.0.
Regards
Petr
[EMAIL PROTECTED]
[EMAIL PROTECTED] napsal dne 25.02.2008 06:18:10:
> Dear R users.
>
> I have written my R code. It runs in the version
>
Hello Gabor and Charilaos
Thanks for the help. I did not realize that the legend argument can
take both the symbol and the line. Following your suggestions I have
come with the code that follows.
I realized that I can use the 'legend' function to draw over ggplot2
produced plot, so that is a good
Course in
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Tartu, Estonia, Wednesday 28 May to Monday 2 June 2008.
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